Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}5x+4y &= 3 \\ 5x-y &= 8\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $5x = y+8$ Divide both sides by $5$ to isolate $x$ $x = {\dfrac{1}{5}y + \dfrac{8}{5}}$ Substitute this expression for $x$ in the first equation. $5({\dfrac{1}{5}y + \dfrac{8}{5}}) + 4y = 3$ $y + 8 + 4y = 3$ Simplify by combining terms, then solve for $y$ $5y + 8 = 3$ $5y = -5$ $y = -1$ Substitute $-1$ for $y$ in the top equation. $5x+4( -1) = 3$ $5x-4 = 3$ $5x = 7$ $x = \dfrac{7}{5}$ The solution is $\enspace x = \dfrac{7}{5}, \enspace y = -1$.